A particle of mass $m_{1}$ is kept at $x = 0$ and another particle of mass $m_{2}$ at $x = d$ . When a third particle is kept at $x = d / 3$ , it experiences no net gravitational force due to the two particles. Find $m_{2} / m_{1}$ .

1 : 4

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4 : 1

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1 : 5

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5 : 1

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Solution

The correct option is B $4 : 1$

The third particle having mass $m (let)$ does not experience net gravitational force due to masses $m_{1}$ and $m_{2}$ .

$∴ \frac{G m_{1} m}{{(d / 3)}^{2}} = \frac{G m_{2} m}{{(d - d / 3)}^{2}}$

$\Rightarrow \frac{9 m_{1}}{d^{2}} = \frac{9 m_{2}}{4 d^{2}}$

$\Rightarrow \frac{m_{2}}{m_{1}} = \frac{4}{1} = 4 : 1$

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A particle of mass m1 is kept at x=0 and another particle of mass m2 at x=d. When a third particle is kept at x=d/3, it experiences no net gravitational force due to the two particles. Find m2/m1.

The correct option is B 4:1 The third particle having mass m (let) does not experience net gravitational force due to masses m1 and m2. ∴Gm1m(d/3)2=Gm2m(d−d/3)2 ⇒9m1d2=9m24d2 ⇒m2m1=41=4:1

A particle of mass $m_{1}$ is kept at $x = 0$ and another particle of mass $m_{2}$ at $x = d$ . When a third particle is kept at $x = d / 3$ , it experiences no net gravitational force due to the two particles. Find $m_{2} / m_{1}$ .